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to any given, but fixed, basis ⁿ.

      The symbol

in the context of properties which are valid independently of the reality or complexity of the inner product.

THEOREM 1.1.– Let (V, 〈 , 〉) be an inner product space. We have:

      1) 〈v, 0_V 〉 = 0 ∀v ∈ V ;

      2) if 〈u, w〉 = 〈v, w〉 ∀w ∈ V , then u and v must coincide;

      3) 〈v, w〉 = 0 ∀v ∈ V

w = 0_V , i.e. the null vector is the only vector which is orthogonal to all of the other vectors.

      PROOF.–

      1) 〈v, 0_V 〉 = 〈v, 0_V + 0_V 〉 = 〈v, 0_V 〉 + 〈v, 0_V 〉 by linearity, i.e. 〈v, 0_V 〉 − 〈v, 0_V 〉 = 0 = 〈v, 0_V 〉.

      2) 〈u, w〉 = 〈v, w〉 ∀w ∈ V implies, by linearity, that 〈u − v, w〉 = 0 ∀w ∈ V and thus, notably, considering w = u − v, we obtain 〈u − v, u − v〉 = 0, implying, due to the definite positiveness of the inner product, that u − v = 0_V , i.e. u = v.

      3) If w = 0_V , then 〈v, w〉 = 0 ∀v ∈ V using property (1). Inversely, by hypothesis, it holds that 〈v, w〉 = 0 = 〈v, 0_V 〉 ∀v ∈ V , but then property (2) implies that w = 0_V .

      Finally, let us consider a typical property of the complex inner product, which results directly from a property of complex numbers.

      THEOREM 1.2.– Let (V, 〈 , 〉) be a complex inner product space. Thus:

      PROOF.– Consider any complex number z = a + ib, so −iz = b − ia, hence b = ℑ (z) = ℜ (−iz). Taking z = 〈v, w〉, we obtain ℑ (〈v, w〉) = ℜ (−i〈v, w〉) = ℜ (〈v, iw〉) by sesquilinearity.

1.2. The norm associated with an inner product and normed vector spaces

      If (V, 〈, 〉) is an inner product space over

, then a norm on V can be defined as follows:

      Note that ‖v‖ is well defined since 〈v, v〉 ≽ 0 ∀v ∈ V . Once a norm has been established, it is always possible to define a distance between two vectors v, w in V : d(v, w) = ‖v − w‖.

The vector v ∈ V such that ‖v‖= 1 is known as a unit vector. Every vector v ∈ V can be normalized to produce a unit vector, simply by dividing it by its norm.

      NOTABLE EXAMPLES.–

      Three properties of the norm, which should already be known, are listed below. Taking any v, w ∈ V , and any α ∈

:

      1) ‖v‖≽ 0, ‖v‖= 0

v = 0_V ;

      2) ‖αv‖= |α|‖v‖(homogeneity);

      3) ‖v + w‖≼ ‖v‖+ ‖w‖(triangle inequality).

      DEFINITION 1.4 (normed vector space).– A normed vector space is a pair (V, ‖ ‖) given by a vector space V and a function, called a norm, , satisfying the three properties listed above.

      A norm ‖ ‖ is Hilbertian if there exists an inner product 〈 , 〉 on V such that .

      Canonically, an inner product space is therefore a normed vector space. Counterexamples can be used to show that the reverse is not generally true.

      Note that, by definition, 〈v, v〉 = ‖v‖ ‖v‖, but, in general, the magnitude of the inner product between two different vectors is dominated by the product of their norms. This is the result of the well-known inequality shown below.

      THEOREM 1.3 (Cauchy-Schwarz inequality).– For all v, w ∈ (V, 〈 , 〉) we have:

      PROOF.– Dozens of proofs of the Cauchy-Schwarz inequality have been produced. One of the most elegant proofs is shown below, followed by the simplest one:

– first proof: if w = 0_V , then the inequality is verified trivially with 0 = 0. If w ≠ 0_V , then we can define , i.e. and we note that:

      thus:

      as the two intermediate terms in the penultimate step are zero, since 〈z, w〉 = 〈w, z〉 = 0.

      As ‖z‖² ≽ 0, we have seen that:

      i.e. |〈v, w〉|² ≼ ‖v‖²‖w‖², hence |〈v, w〉| ≼ ‖v‖‖w‖;

      – second proof (in one line!): ∀t ∈ ℝ we have:

      The Cauchy-Schwarz inequality allows the concept of the angle between two vectors to be generalized for abstract vector spaces. In fact, it implies the existence of a coefficient k between −1 and +1 such that 〈v, w〉 = ‖v‖‖w‖k, but, given that the restriction of cos to [0, π] creates a bijection with [−1, 1], this means that there is only one ϑ ∈ [0, π] such that 〈v, w〉 = ‖v‖‖w‖

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